1. **State the problem:** We have the system of linear equations: $$4x + y + z = 6$$ $$x + 3y + z = 5$$ $$x + y + 2z = 4$$ We need to form the augmented matrix, reduce it to echelon form, and solve using backward substitution. 2. **Form the augmented matrix:** $$\left[\begin{array}{ccc|c}4 & 1 & 1 & 6 \\ 1 & 3 & 1 & 5 \\ 1 & 1 & 2 & 4\end{array}\right]$$ 3. **Reduce to echelon form:** - Use row operations to get zeros below the leading 1 in the first column. - First, swap $R_1$ and $R_2$ to get a leading 1 on top: $$\left[\begin{array}{ccc|c}1 & 3 & 1 & 5 \\ 4 & 1 & 1 & 6 \\ 1 & 1 & 2 & 4\end{array}\right]$$ - Eliminate below $R_1$: $$R_2 = R_2 - 4R_1: (4 - 4\times1, 1 - 4\times3, 1 - 4\times1, 6 - 4\times5) = (0, -11, -3, -14)$$ $$R_3 = R_3 - R_1: (1 - 1, 1 - 3, 2 - 1, 4 - 5) = (0, -2, 1, -1)$$ Matrix now: $$\left[\begin{array}{ccc|c}1 & 3 & 1 & 5 \\ 0 & -11 & -3 & -14 \\ 0 & -2 & 1 & -1\end{array}\right]$$ - Next, eliminate below the leading coefficient in $R_2$: $$R_3 = R_3 - \frac{-2}{-11} R_2 = R_3 - \frac{2}{11} R_2$$ Calculate: $$R_3 = (0, -2, 1, -1) - \frac{2}{11}(0, -11, -3, -14) = (0, -2 + 2, 1 + \frac{6}{11}, -1 + \frac{28}{11}) = (0, 0, \frac{17}{11}, \frac{17}{11})$$ Matrix in echelon form: $$\left[\begin{array}{ccc|c}1 & 3 & 1 & 5 \\ 0 & -11 & -3 & -14 \\ 0 & 0 & \frac{17}{11} & \frac{17}{11}\end{array}\right]$$ 4. **Backward substitution:** - From the last row: $$\frac{17}{11} z = \frac{17}{11} \implies z = 1$$ - From the second row: $$-11 y - 3 z = -14 \implies -11 y - 3(1) = -14 \implies -11 y = -11 \implies y = 1$$ - From the first row: $$x + 3 y + z = 5 \implies x + 3(1) + 1 = 5 \implies x + 4 = 5 \implies x = 1$$ **Final solution:** $$x = 1, y = 1, z = 1$$